Question

A weight Mg is suspended from the middle of a rope whose ends are at the same level. The rope is no longerhorizontal. The minimum tension required to completely straighten the rope is

Answer 1

Draw horizontal and vertical components of Tension.

You will see that Vertical components T sin(Q) + T sin(Q) will balance the weight Mg
Mg = 2T sin(Q)
T = Mg / 2 sin(Q)

--In order to straighten the rope, theta must be 0T = Mg / 2 sin(0) = Mg/0 = infinite-Comon! hang a weight in the middle of a thread and try to straighten it, you won't be able to straighten it. It will always bend a little at the mid point.

Answer 2

Answer:

Minimum Tension, T= Infinitely Large.

Explanation:

[Refer to the attached image 1 to understand the initial case mentioned in the question]

Let the horizontal line represent the rope. Now, Mg weight is suspended from the middle of the rope whose ends are at the same level implying that the tension forces at point mg have been divided equally.

[Refer to attached image 2 to understand the suspension case]

We are asked to find the minimum tension required to completely straighten the rope. Let the tensions make angle ∅ with the horizontal on both sides.

We find the components of Tensions as;-

          Ty = T cos∅ (For both T)

          Tx = T sin∅ (For both T)

Now:-

We know that, when we completely straighten the rope, the angle ∅ will be = 0.

So, in the given case T cos∅ and T cos∅ will result to 0 since they are equal and opposite. We are now left with Mg and 2T sin∅.

[Refer to attached image 3 to understand this case]

     T sin∅ + T sin∅ = Mg

              2T sin∅ = Mg

                    T = $\frac{Mg}{2sin 0}$

                    T = $\frac{Mg}{2}$ × $Cosec0$

                    T = $\frac{Mg}{2}$ × ∞

                    T = Infinitely Large.

Hope it helps! ;-))